Insights Why Do People Say That 1 and .999 Are Equal? - Comments

[Mark44]

I would think this whole theme is answered by:
The definition of real numbers says that $$0.999..._{\text{as real number}} \ =_{\text{acting on real numbers}} \ 1_{\text{as real number}}$$
If you don't like it, use other numbers.

I have no idea what you're trying to say in the equation above.

I think equivalence is more intuitive if you simply state that the limit of 0.999... as we keep adding more decimal places is 1. I think this is also a closer translation of the mathematical statement into plain english.

There's a difference between what you say above and what I think you're trying to say.
First: limit of .999 ... is meaningless, as you aren't saying what is changing.
Second, $\lim_{n \to \infty}\sum_{j = 1}^n \frac 9 {10^j} = 1$. The sum can be made arbitrarily close to 1 just by taking more terms in the sum.

 

[Mark44]

I have no way to say this is true for sure, but it appears to me as though proof #3 is based off an incorrect assumption. Wouldn't the difference between .999... and 1 be infinitesimal, not 0.

There is no need that I can see to bring infinitesimals into the argument. By definition of the notation, .999 ... means an infinitely long string of 9s. If there is a difference between .999 ... and 1, there has to be a digit somewhere in the difference that isn't 0. Which digit is that?

Isn't this one of the reasons we have the concept of infinity, and in this case an infinitesimal value. If something infinitesimal is just considered 0, wouldn't a Riemann sum not mater at all, and just be equal to zero? Wouldn't that make anything using dx, dy, etc. represent a change of zero, as they are defined as an infinitesimaly small change in their respective variable? This is just my thinking, I am not being ignorent, I just want to understand.

 

[WWGD]

I have no way to say this is true for sure, but it appears to me as though proof #3 is based off an incorrect assumption. Wouldn't the difference between .999... and 1 be infinitesimal, not 0. Isn't this one of the reasons we have the concept of infinity, and in this case an infinitesimal value. If something infinitesimal is just considered 0, wouldn't a Riemann sum not mater at all, and just be equal to zero? Wouldn't that make anything using dx, dy, etc. represent a change of zero, as they are defined as an infinitesimaly small change in their respective variable? This is just my thinking, I am not being ignorent, I just want to understand.

Like someone pointed out, there are no infinitesimals in the Standard Real numbers; these exist only on non-standard models of the Reals. Your other questions are good, let me think about them some more. I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0; they become close to 0 in a controlled way. Good point, though.

 

[Mark44]

If something infinitesimal is just considered 0, wouldn't a Riemann sum not mater at all, and just be equal to zero? Wouldn't that make anything using dx, dy, etc. represent a change of zero, as they are defined as an infinitesimaly small change in their respective variable? This is just my thinking, I am not being ignorent, I just want to understand.

In the Riemann definition of a definite integral, you have the number of rectangles n approaching infinity (i.e., increasing without bound), where the area of each rectangle is approaching zero.
In effect, what you have is the indeterminate form $[0 \cdot \infty]$. Indeterminate forms are a way to categorize limits that involve a product or quotient or other operation in which one or more of the quantities is unbounded. Indeterminate forms are not number, and different instances of the same indeterminate form can come out to be literally any real number or $\infty$ or $-\infty$. With Reimann sums, if you have a well-behaved function, the limit comes out to a real number.

Like someone pointed out, there are no infinitesimals in the Standard Real numbers; these exist only on non-standard models of the Reals. Your other questions are good, let me think about them some more. I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0; they become close to 0 in a controlled way. Good point, though.

See above...

 

[WWGD]

In the Riemann definition of a definite integral, you have the number of rectangles n approaching infinity (i.e., increasing without bound), where the area of each rectangle is approaching zero.
In effect, what you have is the indeterminate form $[0 \cdot \infty]$. Indeterminate forms are a way to categorize limits that involve a product or quotient or other operation in which one or more of the quantities is unbounded. Indeterminate forms are not number, and different instances of the same indeterminate form can come out to be literally any real number or $\infty$ or $-\infty$. With Reimann sums, if you have a well-behaved function, the limit comes out to a real number.
See above...

Yes, what I meant is for a specific Riemann sum, the widths are uniformly bounded; the function Min width is a continuous function on a compact set -- integration is done on a compact interval [a,b] -- so it assumes an actual minimum value on [a,b]. Now, we can consider what happens as we let the widths go to 0, but for a _specific_ Riemann sum, the widths do not go to 0. And then we define convergence using nets.

 

[Mark44]

.

Yes, what I meant is for a specific Riemann sum, the widths are uniformly bounded; the function Min width is a continuous function on a compact set -- integration is done on a compact interval [a,b] -- so it assumes an actual minimum value on [a,b]. Now, we can consider what happens as we let the widths go to 0, but for a _specific_ Riemann sum, the widths do not go to 0.

Sure they do. Assuming for simplicity's sake that $\Delta x = \frac{b - a} n$, as $n \to \infty, \Delta x \to 0$. There is no positive minimum value that's a lower bound on $\Delta x.$

For the same reason, we can evaluate $\lim_{x \to \infty} x \cdot \frac 1 {2x + 3}$ and get 1/2, even though there is no upper bound on x, and no positive lower bound on $\frac 1 {2x + 3}$.

And then we define convergence using nets.

 

[WWGD]

.Sure they do. Assuming for simplicity's sake that $\Delta x = \frac{b - a} n$, as $n \to \infty, \Delta x \to 0$. There is no positive minimum value that's a lower bound on $\Delta x.$

For the same reason, we can evaluate $\lim_{x \to \infty} x \cdot \frac 1 {2x + 3}$ and get 1/2, even though there is no upper bound on x, and no positive lower bound on $\frac 1 {2x + 3}$.

Yes, but a specific Riemann sum takes on a specific value for $n$. Then you consider different Riemann sums where n becomes larger. So you have a collection of Riemann sums, each with a specific value of n, and we consider what happens as we consider specific sums with increasingly-larger values for $n$. So you end up with a net of Riemann sums , where the order is inclusion of partition set {#x_1,x_2,...,x_k#} , that converges ( under the right conditions) as a net to the value of the integral.

 

[Mark44]

Yes, but a specific Riemann sum takes on a specific value for $n$. Then you consider different Riemann sums where n becomes larger.

But the Riemann integral is the limit as the partition gets finer, of these sums.
https://en.wikipedia.org/wiki/Riemann_integral

Loosely speaking, the Riemann integral is the limit of the Riemann sums of a function as the partitions get finer.

So you have a collection of Riemann sums, each with a specific value of n, and we consider what happens as we consider specific sums with increasingly-larger values for $n$.

In other words, as you take the limit as n goes to infinity, which forces the subinterval width to zero.

 

[WWGD]

But the Riemann integral is the limit as the partition gets finer, of these sums.
https://en.wikipedia.org/wiki/Riemann_integral

In other words, as you take the limit as n goes to infinity, which forces the subinterval width to zero.

This is kind of difficult to explain in words, but my point is that, for any individual Riemann sum, the function (partition width) is a continuous function on a compact set [a,b], so it assumes a minimum value. Of course, we let the minimal values go to 0.

 

[Mark44]

This is kind of difficult to explain in words, but my point is that, for any individual Riemann sum, the function (partition width) is a continuous function on a compact set [a,b], so it assumes a minimum value. Of course, we let the minimal values go to 0.

A particular Riemann sum will involve a finite number of subintervals of nonzero width. The Riemann integral, however, is the limit of these Riemann sums, as the subinterval width goes to zero.

 

[WWGD]

A particular Riemann sum will involve a finite number of subintervals of nonzero width. The Riemann integral, however, is the limit of these Riemann sums, as the subinterval width goes to zero.

So we are saying the same thing. I was addressing a poster question about the fact that an indefinitely-small Standard Real number must equal 0, which led them to believe that it must follow that a Riemann sum must equal 0 , since all the terms in the sum become indefinitely small. But this is not so, because in an individual/specific Riemann sum, the terms do not become indefinitely small; they are uniformly bounded by my previous argument. Of course this bound goes to 0, but the minimal width on a specific Riemann sum does not go to 0 or else the sum itself would be zero. Now, it would be interesting how to do integrals in non-standard analysis. I have seen methods to differentiate but I have not seen integration theory in non-standard analysis.

 

[Mark44]

I'm not sure that we were. Isaac0427 said

wouldn't a Riemann sum not mater at all, and just be equal to zero?

I believe he really meant Riemann integral here.

Your response to him was

I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0; they become close to 0 in a controlled way.

Your explanation used a lot of fairly high-powered analysis, but was flawed; i.e., that the (areas of) the rectangles are going to zero while their widths miraculously remain positive, or that they "become close to zero in a controlled way."

 

[WWGD]

I'm not sure that we were. Isaac0427 said I believe he really meant Riemann integral here.

Your response to him was
Your explanation used a lot of fairly high-powered analysis, but was flawed; i.e., that the (areas of) the rectangles are going to zero while their widths miraculously remain positive, or that they "become close to zero in a controlled way."

Do you see something specifically wrong with the claim that rectangle width is a continuous function on a compact set so it achieves a minimum, which is a uniform lower bound? Is this _not_ a controlled way? What flaw do you see here? Or do you claim that in a specific Riemann sum the widths, and specifically the minimal width, becomes indefinitely close to 0. Convince me then that the integral is not zero.

 

[Mark44]

Do you see something specifically wrong with the claim that rectangle width is continuous on a compact set so it achieves a minimum, which is a uniform lower bound? Is this _not_ a controlled way? What flaw do you see here?

This part, which I just quoted:

so these widths do not become indefinitely-close to 0;

 

[WWGD]

This part, which I just quoted:

In a _specific sum_ , the function interval width is continuous and defined on a compact set. What is wrong with this argument? If you let your minimal width go to 0 in a _specific sum_ without bound, your integral will equal 0. What you end up getting is a net of Riemann sums, which will converge netwise to the value of the integral under the right conditions.

 

[WWGD]

I also see it like this: The collection {$x_1=a, x_2,..., x_n=b$} covers [a,b]. By compactness of [a,b], the collection has a finite subcover. Now, consider min $(x_{i+1}-x_i )$:=m on the finite subcover. A finite set will have a minimum value , say m. The value $m=0$ will never be realized. Maybe I am not explaining myself well here.

 

[WWGD]

Sorry, could not edit. Just to say I need to get to sleep, I will address any other disagreement tomorrow.

 

[Mark44]

In a _specific sum_ , the function interval width is continuous and defined on a compact set. What is wrong with this argument? If you let your minimal width go to 0 in a _specific sum_ without bound, your integral will equal 0. What you end up getting is a net of Riemann sums, which will converge netwise to the value of the integral under the right conditions.

We are not talking about a specific Riemann sum. My complaint is that you are talking about a limit, but don't seem to realize it. Emphasis added.

I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0

In other words, as $\Delta x \to 0$ -- that's a limit. All the other stuff about compact sets, continuity, and subcovers etc. is just word salad that I'm 99% sure would be over the head of the person you were responding to.

 

[madness]

There's a difference between what you say above and what I think you're trying to say.
First: limit of .999 ... is meaningless, as you aren't saying what is changing.
Second, $\lim_{n \to \infty}\sum_{j = 1}^n \frac 9 {10^j} = 1$. The sum can be made arbitrarily close to 1 just by taking more terms in the sum.

But what I actually said was "limit of 0.999... as you keep on adding more decimal places", which is not the same as what you quoted from me. I'm not sure what point you are making in the second part, as it seems to agree with what I wrote. The mathematical statement simply says that as you make n bigger and bigger you get closer and closer to 1, which I don't think anybody would disagree with.

 

[PeroK]

But what I actually said was "limit of 0.999... as you keep on adding more decimal places", which is not the same as what you quoted from me. I'm not sure what point you are making in the second part, as it seems to agree with what I wrote. The mathematical statement simply says that as you make n bigger and bigger you get closer and closer to 1, which I don't think anybody would disagree with.

It is important, though, to realize that a limit is a number, and not a process. It's a small point, perhaps, but it sharpens your understanding of these things.

 

[madness]

It is important, though, to realize that a limit is a number, and not a process. It's a small point, perhaps, but it sharpens your understanding of these things.

Yes, but it's also important to remember that this number (the limit) is the number that you get closer and closer to as you apply the process of increasing n. Or, if you don't like this process, it's the number that you can get as close to as you like by choosing a high enough value of n.

 

[PeroK]

Yes, but it's also important to remember that this number (the limit) is the number that you get closer and closer to as you apply the process of increasing n. Or, if you don't like this process, it's the number that you can get as close to as you like by choosing a high enough value of n.

Neither of those is the definition of a limit.

 

[HallsofIvy]

Neither of those is the definition of a limit.

Neither is a rigorous definition of a limit but they are a statement of what is implied by such a definition. In particular, the limit of the sequence 0.9, 0.99, 0.999, ... is the number 1, not the sequence itself.

 

[PeroK]

Neither is a rigorous definition of a limit but they are a statement of what is implied by such a definition. In particular, the limit of the sequence 0.9, 0.99, 0.999, ... is the number 1, not the sequence itself.

Perhaps, but my original point was that there are those who see the limit as a number (full stop) and those who see the limit as the hypothetical result of a never-ending process. Breaking away from this latter view is perhaps an important step in grasping rigorous maths.

 

[jbriggs444]

Perhaps, but my original point was that there are those who see the limit as a number (full stop) and those who see the limit as the hypothetical result of a never-ending process. Breaking away from this latter view is perhaps an important step in grasping rigorous maths.

The phrasing that madness used was careful and avoided speaking of the result of a never-ending process.

Yes, but it's also important to remember that this number (the limit) is the number that you get closer and closer to as you apply the process of increasing n. Or, if you don't like this process, it's the number that you can get as close to as you like by choosing a high enough value of n.

 

[PeroK]

The phrasing that madness used was careful and avoided speaking of the result of a never-ending process.

Fair enough. It's not the way I read it, but my apologies to madness.

 

[WWGD]

We are not talking about a specific Riemann sum. My complaint is that you are talking about a limit, but don't seem to realize it. Emphasis added.

In other words, as $\Delta x \to 0$ -- that's a limit. All the other stuff about compact sets, continuity, and subcovers etc. is just word salad that I'm 99% sure would be over the head of the person you were responding to.

No, I think you are not realizing my point. I am talking about an individua Riemann sum not being 0, addressing the question. I do know that the Riemann sum is the limit ( if it exists) of the Riemann sums. I never said nor implied otherwise. I did reply to the OP that for individual sums, the width is controlled, so it does not go to zero. The rest of the exchanges were with and for you.

 

[Mark44]

No, I think you are not realizing my point. I am talking about an individua Riemann sum not being 0, addressing the question. I do know that the Riemann sum is the limit ( if it exists) of the Riemann sums.

I think you mean Riemann integral being the limit of the Riemann sums.

I never said nor implied otherwise. I did reply to the OP that for individual sums, the width is controlled, so it does not go to zero. The rest of the exchanges were with and for you.

My complaint is in the following. I interpret the phrase "while going to 0" as equivalent to saying "the limit of the Riemann sums as n goes to infinity." You can't say that the widths of the rectangles are going to zero, but at the same time do not become close to zero. Here again is what you said, with emphasis added.

I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0

The rectangle widths are either going to zero or they're not -- you can't have it both ways.

 

[WWGD]

I think you mean Riemann integral being the limit of the Riemann sums.My complaint is in the following. I interpret the phrase "while going to 0" as equivalent to saying "the limit of the Riemann sums as n goes to infinity." You can't say that the widths of the rectangles are going to zero, but at the same time do not become close to zero. Here again is what you said, with emphasis added.

The rectangle widths are either going to zero or they're not -- you can't have it both ways.

Yes, I meant Riemann integral, my bad. I was referring to individual Riemann sums, not to the Riemann integral itself. I was addressing the question of whether a Riemann sum is zero or not, because the general term goes to zero in an uncontrolled way. I made no reference to the Riemann integral itself; that is a whole other topic. But yes, the Riemann integral is the limit as the partition width goes to zero. I don't see why the two contradict or are incompatible with each other.

 

[Mark44]

I was referring to individual Riemann sums, not to the Riemann integral itself. I was addressing the question of whether a Riemann sum is zero or not, because the general term goes to zero in an uncontrolled way.

When you say, "the rectangles, while going to 0, ..." you're no longer talking about a specific Riemann sum -- you're talking about the limit of the Riemann sums (as the number of terms goes to infinity), which means you're talking about the Riemann integral.

These were your exact words:

I think the point is that the rectangles, while going to 0, are uniformly bounded , by compactness of the region of integration, so these widths do not become indefinitely-close to 0

In short, you're saying that the rectangles are going to zero and that their widths are not going to zero. That's impossible.

Also, I don't know what you mean "goes to zero in an uncontrolled way" ...

I made no reference to the Riemann integral itself; that is a whole other topic. But yes, the Riemann integral is the limit as the partition width goes to zero. I don't see why the two contradict or are incompatible with each other.